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Question Summary: A tetraoxosulphate (VI) salt, P, is heated with sodium hydroxide. A gas, R, is evolved and it turns red litmus paper blue.
---
(a) Identify R:
The gas evolved turns red litmus blue, indicating it is basic. When tetraoxosulphate (VI) salt (like ammonium sulphate) is heated with NaOH, ammonia (NH₃) gas is released.
Answer: R is Ammonia (NH₃).
---
(b) Give the confirmatory test for R:
A common confirmatory test for ammonia gas is:
Bring a glass rod dipped in concentrated hydrochloric acid near the gas. Dense white fumes of ammonium chloride confirm the presence of ammonia.
Answer: Pass the gas over concentrated HCl; white fumes of NH₄Cl confirm ammonia.
---
(c) Write the formula of P:
Since the salt of tetraoxosulphate (VI) is involved and ammonia is released, the salt is likely ammonium tetraoxosulphate (VI).
Answer: The formula of P is (NH₄)₂SO₄.
---
(d) Name P:
Answer: P is Ammonium tetraoxosulphate (VI) or simply Ammonium sulphate.
---Question Summary: A tetraoxosulphate (VI) salt, P, is heated with sodium hydroxide. A gas, R, is evolved and it turns red litmus paper blue.
---
(a) Identify R:
The gas evolved turns red litmus blue, indicating it is basic. When tetraoxosulphate (VI) salt (like ammonium sulphate) is heated with NaOH, ammonia (NH₃) gas is released.
Answer: R is Ammonia (NH₃).
---
(b) Give the confirmatory test for R:
A common confirmatory test for ammonia gas is:
Bring a glass rod dipped in concentrated hydrochloric acid near the gas. Dense white fumes of ammonium chloride confirm the presence of ammonia.
Answer: Pass the gas over concentrated HCl; white fumes of NH₄Cl confirm ammonia.
---
(c) Write the formula of P:
Since the salt of tetraoxosulphate (VI) is involved and ammonia is released, the salt is likely ammonium tetraoxosulphate (VI).
Answer: The formula of P is (NH₄)₂SO₄.
---
(d) Name P:
Answer: P is Ammonium tetraoxosulphate (VI) or simply Ammonium sulphate.
---
(2a)
Procedure:
Add all of C into a beaker with 10cm³ of distilled water and stir. Test the resulting solution with both red and blue litmus papers.
Observation:
Blue litmus paper turns red, red litmus paper remains unchanged.
Inference:
The solution is acidic.
(2bi)
First portion: Add NaOH(aq) in drops, then in excess
Observation:
A white gelatinous precipitate forms on addition of NaOH, which dissolves in excess NaOH.
Inference:
Presence of aluminium ions (Al³⁺).
(2bii)
Second portion: Add NH₃(aq) in drops, then in excess
Observation:
A white gelatinous precipitate forms, which is insoluble in excess ammonia.
Inference:
Confirms Al³⁺ ion, since Al(OH)₃ does not dissolve in excess NH₃.
(2biii)
Third portion: Add BaCl₂(aq), followed by dilute HCl
Observation:
A white precipitate forms with BaCl₂, which is insoluble in dilute HCl.
Inference:
Indicates the presence of sulfate ions (SO₄²⁻).
(2biv)
Fourth portion: Add Fehling’s solution A and B and boil.
Observation:
No red precipitate forms.
Inference:
No reducing sugar present.
Conclusion:
From all tests, the two substances present in mixture C are:
(i) Aluminium salt (e.g. aluminium sulfate Al₂(SO₄)3) gives Al³⁺ and SO₄²⁻
(ii) No reducing sugar or organic acid detected
